Question

answer this pls............its urgent​

Answer 1

hey mate here is ur solution
Given
a+2d+a+6d=6
2a+8d=6
2(a+4d)=2(3)
a=3-4d......(1)
also given
(a+2d)(a+6d)=8
${a}^{2} + 6ad + 2ad + 12 {d}^{2} = 8\\ {a}^{2} + 8ad + 12 {d}^{2} = 8 \\ put \: the \: value \: of \: a \: in \: the \: eq \\ {(3 - 4d)}^{2} + 8(3 - 4d) d+ 12 {d}^{2} = 8 \\ 9 + 16 {d}^{2} - 24d \: + 24d - 32 {d}^{2} + 12 {d}^{2} = 8 \\ 9 - 8 - 32 {d}^{2} + 28 {d}^{2} = 0 \\ 1 - 4 {d}^{2} = 0 \\ (1 + 4d)(1 - 4d) = 0 \\ d = - \frac{1}{4} or \frac{1}{4}$
put the value of d in eq(1)
$a = 3 - 4d \\ = 3 - 4( - \frac{ 1}{4}) or \: 3 - 4( \frac{1}{4} )\\ = 4 \: or \: 2$
Hope it helps u